NJC H2 Chem 2012 Prelim P2 Soln

NJC H2 Chemistry Prelim Solutions Solutions to Paper 2 1 (a) (i) LiOH NaOH KOH RbOH CsOH · [1/2] · Size of cations increases down Group II, heat of hydration of cation less exothermic [1/2] Most students can score these two marks. The students who drew an increasing line above the x-axis were awarded one upon two. Common mistakes · It is clearly stated in the question that all of the heats of solution are exothermic. Many students did not seem to notice that, judging from their answers. Their lines were above the x-axis. · It is ∑nΔH(hyd), not just nΔH(hyd)! · A lot of students, while getting the full credit for the graph, failed to include the equation in their explanation. (b) The group I hydroxides are all hygroscopic. [1/2] Weigh the solid out promptly so as to minimise its exposure to air. [1/2] (c) Wear gloves. [1/2] The chemicals are corrosive. [1/2] Common mistakes · Toxicity is not the same as corrosiveness! · Grammar! A pair of gloves, not just a glove! If you were to use just one glove, won’t your other hand get corroded? A pair of google s, not just a goggle! If you were to use just one goggle, how are you going to wear it? Hold it over your eye? (d) Procedure Major points [1/2 each, maximum 2] · Measure the initial temperature of the temperature of the water. · Measure the highest/maximum temperature reached. (No marks if temperature is lowest/minimum; ECF according to answer to (a)). · Measuring cylinder to measure out the water. · Weigh mass of the residue and weighing bottle after adding the LiOH into the cup. Minor points [1/2 each, maximum 1] · Weigh out LiOH using a weighing bottle. · Cup to be supported with a beaker. · Repeat the experiment until two values are within 5 % of each other. ‘m’ has to be specified as the mass of the solid. 1

NJC H2 Chemistry Prelim Solutions 2 Preliminary calculations [1] · Justify mass of LiOH weighed out. [NOTE: The capacity of the cup must be stated, otherwise at most half a mark will be awarded.] [Set ] where m is between 50 to 90 % of the stated capacity of cup. Simplifying, If the capacity of the cup is 100 cm3 and filling 90 % of it, then (e) Volumes of the two solutions are the same/same setup/same apparatus [1] (f) (i) · for both experiments. (Check: and ) · no. of moles of H2SO4 half that of CH3COOH. · total volume not more than 90% of capacity of cup and not less than 50% of capacity of cup. · acids are the limiting reagents i.e. C1V1 < 60 and C2V2 < 30. 2 (a)(i) n(O2) = n(S2O32-) = ( ) = [2] [O2] = ( ) = moldm-3 [1/2] [O2] = ( ) = 7.44 mgdm-3 [1/2] (a)(ii) The carps will be able to survive. (Justification).

NJC H2 Chemistry Prelim Solutions 3 (b)(i) (ii) 2H+ + 3SO3 2- + 2ReO4 - H2O + 3SO4 2- + 2ReO2 [1] Ecell0 = (0.3 – 0.17) V = 0.13 V > 0 [1] Dark solid forms. [1] 3(a) Copper, brass and bronze all have metallic bonds. Copper is a pure metal made up of Cu2+ ions (0.069nm) arranged in a orderly metallic lattice. Brass and bronze are alloys and the presence of metal ions of different sizes disrupts the orderly metallic lattice. Ionic radius of Sn2+ ion (0.112nm) is bigger than ionic radius of Zn2+ ions (0.074nm) and the bigger ions Sn2+ will disrupt the orderly metallic lattice more significantly. The greater extent of disruption of the orderly lattice makes it more difficult for the layers of atoms to slide over each other, leading to greater hardness. Or Copper gives out 1 electron to the sea of electron, zinc gives out 2 while tin gives out 4. Ionic radius of Cu+ is approximately 0.069nm, Zn2+ is 0.074nm and Sn4+ is approximately 0.112nm. Charge/size of Cu+ is lowest, followed by Zn2+ and Sn4+. The metallic bond is weakest in Cu followed by Zn and Sn. Therefore, greater force is needed for the layers of atom to slide over each other in Brass and bronze. 3(b) Zinc does not form any ions that have partially filled 3d orbital. 3(c) (i) Type of reaction: Disproportionation Cu + Cu2+ + e E°ox = –0.15V Cu+ + e Cu E°red = +0.52V E°cell = +0.52–0.15 = 0.37V > 0, reaction is feasible.

NJC H2 Chemistry Prelim Solutions 4 (ii) F: [Cu(RNH2)4]2+ G: [CuCl4]2– (iii) [CuCl4]2– + 6H2O [Cu(H2O)6]2+ + 4Cl– (iv) Cream precipitate in brown solution. 2Cu2+ + 4I– 2CuI + I2 (v) Cu2+ has partially filled 3d orbitals. When ligands approach, the 3d orbitals of Cu2+ are splitted into 2 different energy levels. When an electron is promoted from lower energy 3d orbital to higher energy 3d orbital, an energy corresponding to the wavelength in the visible region is absorbed. The complementary colour of the wavelength absorbed is blue. (d) When copper is exposed to air, it is oxidized to CuO. When CuO reacts with the moisture in the air, Cu(OH)2 is formed. Some of the Cu(OH)2 reacts with CO2 in the air to form CuCO3. This leads to the formation of Cu(OH)2.CuCO3. or When copper is exposed to air, it is oxidized to CuO. Some of the CuO reacts with the moisture in the air forming Cu(OH)2 while some of the CuO reacts with CO2 in the air forming CuCO3. This leads to the formation of Cu(OH)2.CuCO3. 4(a) (i) E2: –O2CCH2CH(NH3+)CO2– E3: –O2CCH2CH(NH2)CO2– (ii) pKa1 = 2.11 Ka1 = 10–2.11 = 7.76 x 10–3 (iii) E1: The solution present at E1 can act as a buffer and the pH change is relatively small when small amount of NaOH(aq) is added. E3: The solution present at E3 is a salt and when small amount of NaOH(aq) is added, there is a large increase in pH due to OH– from NaOH. Or E1: At E1, the species still have H+ available to neutralize OH–, however, at E3, the species present does not have any H+ to neutralize OH– from NaOH, hence, the large increase in pH is due to the OH– from NaOH. (iv) the two species present are HO2CCH2CH(NH3+)CO2– and –O2CCH2CH(NH3+)CO2– Salt: –O2CCH2CH(NH3+)CO2– Acid: HO2CCH2CH(NH3+)CO2– pH = pKa – lg( ]acid[ ]salt[ ) 4 =3.86 – lg( ]acid[ ]salt[ ) ]acid[ ]salt[ ) = 1.38 (b) (i) Active site (ii) When pH is low, the concentration of H + from the solution is high, –COO– of asp will be protonated to form –COOH. Therefore, no lone pair is available to abstract H+ from his and subsequent electron

NJC H2 Chemistry Prelim Solutions 5 transfer cannot happen. The will not be –O:– on ser to act as a nucleophile to attack the δ+ C of the peptide bond to bring about hydrolysis. 5(i) 1. Primary amine 2. Tertiary alcohol 3. amide 4. ester (ii) cold HCl(aq) C CH2NH3 + CH2 OH N C CH2CH3 H O OC O C H3 Hot NaOH(aq) CH3CO2Na CH3CH2CO2Na C CH2NH2 CH2 OH NH2NaO 6 (a) (i) Limited Br2 or excess alkane and ultraviolet light/ sunlight/ heat (ii) Free radical substitution [1] Initiation Br Br uv 2 Br Propagation CH3 C H C H CH3 HCH3 C H C H CH3 H H + Br + HBr CH3 C H C H CH3 HCH3 C H C H CH3 H + Br2 Br + Br Termination

NJC H2 Chemistry Prelim Solutions 6 CH3C H C H CH3 CH3 C H C H CH3 CH3 C H C H CH3 H H 2 Br Br2 H 2 (iii) There are 9 primary H atoms available for substitution to form 1-bromo-2-methylpropane compared to 4 secondary H atoms to form 2-bromo-2-methylpropane.By equal probability of substitution, expected ratio of 1-bromo-2-methylpropane : 2-bromo-2-methylpropane is 9 : 1 However, a tertiary radical, •C(CH3)3, is more stable than a primary radical, •CH2CH(CH3)2 due to the electro-donating effects of the alkyl groups. More tertiary radicals are formed, giving rise to more 2-bromo-2-methylpropane formed. Thus the observed percentages. (b) (i) AlBr3 or FeBr3 or AlCl3 or FeCl3 , anhydrous (ii) The Lewis acid catalyst is required to form a strong electrophile (CH 3)2CHCH2+ [1] to react with the aromatic /resonance stabilised benzene. AlBr3 use a vacant orbital of Al to accept/extract a lone pair/ Br‒. [1] OR AlBr3 + R-Br à AlBr4‒ + R+ [1] OR AlCl3 or AlBr3 will dissolve in water and be hydrolys